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easyBasicsSometimes asked

Binary to Decimal

Basics · 6 min

Given a binary string s (e.g. "1011"), return its decimal value as an integer.

In plain English

Process digits left to right using Horner's method: result = result*2 + bit.

Examples

Example 1

Input: s = "101"
Output: 5
Note: 4+1.

Example 2

Input: s = "1000"
Output: 8
Note: 2³.

Example 3

Input: s = "0"
Output: 0
Note: Zero.

Constraints

1 ≤ s.length ≤ 31
s consists of '0' and '1'

Clarifying questions1First question: Could the value overflow?Explore questions

Q1.Could the value overflow?
For long strings I'd use long/BigInteger; here it fits in int.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Start at 0; for each char, do `result = result * 2 + (c - '0')`. O(length).

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you use Horner's method instead of per-digit powers.

Where candidates slip

Power-per-digit, or using the raw char code.

How to stand out

Note it's the same scheme as decimal parsing, generalising to any base.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Horner's method

TimeO(length)SpaceO(1)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Process bits left to right — how does the running value change per bit?

2Hint 2Explore hint

result = result * 2 + bit avoids explicit powers of two.

3Hint 3Explore hint

Convert the char '0'/'1' to its int value.

A · Core idea

Convert a binary string to its integer value with Horner's method: scan left to right, and for each bit do result = result * 2 + bit.

Why Horner's rule works. Reading MSB-first, each new bit means the existing value shifts left one place (×2, like appending a binary digit) and the new bit is added in the units place. This evaluates the polynomial b₀·2ᵏ + … + bₖ without ever computing explicit powers of two — it's the same scheme as parsing a decimal number, just base 2.

Why not pow(2, i) per digit. Computing a power for each position is slower and more overflow-prone; Horner's needs only one multiply-add per bit.

Char-to-int. Convert each character with c - '0' to get 0 or 1; forgetting this and using the char's ASCII code is a common slip.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

s = "101"

'1'

0*2 + 1

Stateresult=1

Final result5

C · Code with teaching comments

solution.javaJava

Horner's method — optimal

1 teaching note

1
public int toDecimal(String s) {
2
    int result = 0;
3
    for (int i = 0; i < s.length(); i++) {
4
        // Double the accumulator and add the new bit.
5
        result = result * 2 + (s.charAt(i) - '0');
6
    }
7
    return result;
8
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.8 lines

D · Why it works and complexityO(length) timeO(1) spaceOpen the invariant and proofExplore the proof

ComplexityO(length) time, O(1) space — one multiply-add per character — linear in the string length.

E · Edge cases and pitfalls5s = "1000" — → 8.Explore edge cases

Edge cases handled

s = "1000"→ 8.
s = "0"→ 0.
s = "1111"→ 15.

Pitfalls

Computing pow(2, i) per digit (slower, overflow-prone).
Not converting the char '0'/'1' to its int value.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic int toDecimal(String s) { int result = 0; for (int i = 0; i < s.length(); i++) { result = result * 2 + (s.charAt(i) - '0'); }Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Computing pow(2, i) per digit (slower, overflow-prone).
Not converting the char '0'/'1' to its int value.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Horner's method

Code pattern to memorize

result = 0; for c: result = result*2 + (c-'0')

Rules to remember

1result = result*2 + bit.
2Left to right.
3char - '0' for the digit.

Use this pattern when

Parsing numbers in any base

Interview takeaway

Horner's rule turns digit parsing into one multiply-add per char.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Horner's method. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Decimal to Binary
The inverse via repeated division.

Same pattern

easyBasicsSometimes asked

Binary to Decimal

Basics · 6 min

Given a binary string s (e.g. "1011"), return its decimal value as an integer.

In plain English

Process digits left to right using Horner's method: result = result*2 + bit.

Examples

Example 1

Input: s = "101"
Output: 5
Note: 4+1.

Example 2

Input: s = "1000"
Output: 8
Note: 2³.

Example 3

Input: s = "0"
Output: 0
Note: Zero.

Constraints

1 ≤ s.length ≤ 31
s consists of '0' and '1'

Clarifying questions1First question: Could the value overflow?Explore questions

Q1.Could the value overflow?
For long strings I'd use long/BigInteger; here it fits in int.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Start at 0; for each char, do `result = result * 2 + (c - '0')`. O(length).

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you use Horner's method instead of per-digit powers.

Where candidates slip

Power-per-digit, or using the raw char code.

How to stand out

Note it's the same scheme as decimal parsing, generalising to any base.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Horner's method

TimeO(length)SpaceO(1)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Process bits left to right — how does the running value change per bit?

2Hint 2Explore hint

result = result * 2 + bit avoids explicit powers of two.

3Hint 3Explore hint

Convert the char '0'/'1' to its int value.

A · Core idea

Convert a binary string to its integer value with Horner's method: scan left to right, and for each bit do result = result * 2 + bit.

Why not pow(2, i) per digit. Computing a power for each position is slower and more overflow-prone; Horner's needs only one multiply-add per bit.

Char-to-int. Convert each character with c - '0' to get 0 or 1; forgetting this and using the char's ASCII code is a common slip.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

s = "101"

'1'

0*2 + 1

Stateresult=1

Final result5

C · Code with teaching comments

solution.javaJava

Horner's method — optimal

1 teaching note

1
public int toDecimal(String s) {
2
    int result = 0;
3
    for (int i = 0; i < s.length(); i++) {
4
        // Double the accumulator and add the new bit.
5
        result = result * 2 + (s.charAt(i) - '0');
6
    }
7
    return result;
8
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.8 lines

D · Why it works and complexityO(length) timeO(1) spaceOpen the invariant and proofExplore the proof

ComplexityO(length) time, O(1) space — one multiply-add per character — linear in the string length.

E · Edge cases and pitfalls5s = "1000" — → 8.Explore edge cases

Edge cases handled

s = "1000"→ 8.
s = "0"→ 0.
s = "1111"→ 15.

Pitfalls

Computing pow(2, i) per digit (slower, overflow-prone).
Not converting the char '0'/'1' to its int value.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic int toDecimal(String s) { int result = 0; for (int i = 0; i < s.length(); i++) { result = result * 2 + (s.charAt(i) - '0'); }Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Computing pow(2, i) per digit (slower, overflow-prone).
Not converting the char '0'/'1' to its int value.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Horner's method

Code pattern to memorize

result = 0; for c: result = result*2 + (c-'0')

Rules to remember

1result = result*2 + bit.
2Left to right.
3char - '0' for the digit.

Use this pattern when

Parsing numbers in any base

Interview takeaway

Horner's rule turns digit parsing into one multiply-add per char.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Horner's method. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Decimal to Binary
The inverse via repeated division.

Same pattern